Find the vertical asymtone:

f(x)=`(x-2)/(2x^(2)+3x-2)`

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To determine the value(s) of x for which the denominator is 0, solve `2x^2+3x-2=0`. Use the quadratic formula:

`x=(-b+-sqrt(b^2-4ac))/(2a)` for an equation of the form `ax^2+bx+c=0`

Substitute 3 for b, 2 for a and -2 for c.

`x=(-3+-sqrt((3)^2-(4)(2)(-2)))/((2)(2))`

`x=(-3+-sqrt(9+16))/4`

`x=(-3+-5)/4`

`x=-8/4=-2`

`x=2/4=0.5`

**The function is undefined for x=-2 and x=0.5, thus x=-2 and x=0.5 are the asymptotes.**

Rational functions (functions with a ratio of polynomials) often have vertical asymptotes when the denominator equals zero.

In this case, we can factor the denominator:

`2x^2 + 3x - 2 = (2x-1)(x+2)`

Therefore the denominator equals zero AND we have vertical asymptotes when

`x = 1/2`

`x=-2`

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